Question

If P is pressure, $\rho$ is the density then $\frac{P}{\rho }$ has the same dimensions of :

• A. Force per unit mass
• B. Energy per unit mass
• C. Power per unit velocity
• D. Relative density

Answer 1

The correct option is B Energy per unit mass

$P=pressure=\frac{Force}{Area}=\frac{M^1{L}^1{T}^{-2}}{L^2}$$=\left[M^1{L}^{-1}{T}^{-2}\right]$
$\rho =\left[M^1{L}^{-3}{T}^0\right]$

$\frac{P}{\rho }=\frac{M^1{L}^{-1}{T}^{-2}}{M^1{L}^{-3}}$$=\left[M^0{L}^2{T}^{-2}\right]$

Force per unit mass $=\frac{M^1{L}^1{T}^{-2}}{M^1}=\left[L^1{T}^{-2}\right]$
Energy per unit mass $=\frac{M^1{L}^2{T}^{-2}}{M^1}=\left[L^2{T}^{-2}\right]$
Power per unit velocity $=\frac{M^1{L}^2{T}^{-3}}{L{T}^1}=\left[M^1{L}^1{T}^{-2}\right]$
Relative density $=\left[M^0{L}^0{T}^0\right]$

Hence, option B is correct.